Rearrange the kinetic energy equation so that velocity is the subject
1 answer:
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Answer:
the magnitude of Vpg = 493.711 km/h
Explanation:
given data
speed Vpg = 560 km/h
speed Vwg = 80 km/h
solution
we get here magnitude of the plane velocity w.r.t. ground is
we know that the Vpg = Vpw + Vwg .....................1
writing the component of the velocity that is
Vpw = (0 km/h î - 560 km/h j )
Vwg = (80 cos 45 km/h î + 80 sin 45 km/h j)
adding these
Vpg = (0+80 cos 45 km/h ) î + ( -560 + 80 sin 45 km/h j)i
Vpg = (42.025 ) î (-491.92 km/h)j
now we take magnitude
the magnitude of Vpg =
the magnitude of Vpg = 493.711 km/h
Answer:
v = 6.195 m / s
Explanation:
For this exercise we can use the conservation of energy, for the system formed by the ball and the pulley
starting point. Higher before releasing the system
Em₀ = U = M g h
final point. When the ball has lowered h = 2
Em_f = K = ½ M v² + ½ I w²
the energy is preserved
Em₀ = Em_f
M g h = ½ M v² + ½ I w²
angular and linear velocity are related
v = w r
w = v / r
indicate that the moment of inertia is
I = ½ m r²
we substitute
M g h = ½ M v² + ½ (½ m r²) (v/r) ²
½ v² (M + m) = M g h
v² =
let's calculate
v =
v = 6.195 m / s